Calculus/Mean Value Theorem

If f(x){\displaystyle f(x)} is continuous on the closed interval [a,b]{\displaystyle [a,b]} and differentiable on the open interval (a,b){\displaystyle (a,b)} , there exists a number c∈(a,b){\displaystyle c\in (a,b)} such that

f′(c)=f(b)−f(a)b−a{\displaystyle f'(c)={\frac {f(b)-f(a)}{b-a}}}

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What does this mean? As usual, let us utilize an example to grasp the concept. Visualize (or graph) the function f(x)=x3{\displaystyle f(x)=x^{3}} . Choose an interval (anything will work), but for the sake of simplicity, [0,2]. Draw a line going from point (0,0) to (2,8). Between the points x=0{\displaystyle x=0} and x=2{\displaystyle x=2} exists a number x=c{\displaystyle x=c} , where the derivative of f{\displaystyle f} at point c{\displaystyle c} is equal to the slope of the line you drew.

2: By the definition of the mean value theorem, we know that somewhere in the interval exists a point that has the same slope as that point. Thus, let us take the derivative to find this point x=c{\displaystyle x=c} .

dydx=3x2{\displaystyle {\frac {dy}{dx}}=3x^{2}}

Now, we know that the slope of the point is 4. So, the derivative at this point c{\displaystyle c} is 4. Thus, 4=3x2{\displaystyle 4=3x^{2}} . The square root of 4/3 is the point.

Example 2: Find the point that satisifes the mean value theorem on the function f(x)=sin⁡(x){\displaystyle f(x)=\sin(x)} and the interval [0,π]{\displaystyle [0,\pi ]} .

(Remember, sin⁡(π){\displaystyle \sin(\pi )} and sin⁡(0){\displaystyle \sin(0)} are both 0.)

2: Now that we have the slope of the line, we must find the point x=c{\displaystyle x=c} that has the same slope. We must now get the derivative!

dsin⁡(x)dx=cos⁡(x)=0{\displaystyle {\frac {d\sin(x)}{dx}}=\cos(x)=0}

The cosine function is 0 at π2+kπ{\displaystyle {\frac {\pi }{2}}+k\pi } , where k{\displaystyle k} is an integer. Remember, we are bound by the interval [0,π]{\displaystyle [0,\pi ]} , so π2{\displaystyle {\frac {\pi }{2}}} is the point c{\displaystyle c} that satisfies the Mean Value Theorem.

The "Differential of x{\displaystyle x}" is the Δx{\displaystyle \Delta x} . This is an approximate change in x{\displaystyle x} and can be considered "equivalent" to dx{\displaystyle dx} . The same holds true for y{\displaystyle y} . What is this saying? One can approximate a change in y{\displaystyle y} by knowing a change in x{\displaystyle x} and a change in x{\displaystyle x} at a point very nearby. Let us view an example.

Example: A schoolteacher has asked her students to discover what 4.12{\displaystyle 4.1^{2}} is. The students, bereft of their calculators, are too lazy to multiply this out by hand or in their head and desire to utilize calculus. How can they approximate this?

1: Set up a function that mimics the procedure. What are they doing? They are taking a number (Call it x{\displaystyle x}) and they are squaring it to get a new number (call it y{\displaystyle y}). Thus, y=x2{\displaystyle y=x^{2}} Write yourself a small chart. Make notes of values for x,y,Δx,Δy,dydx{\displaystyle x,y,\Delta x,\Delta y,{\frac {dy}{dx}}} . We are seeking what y{\displaystyle y} really is, but we need the change in y{\displaystyle y} first.

2: Choose a number close by that is easy to work with. Four is very close to 4.1, so write that down as x{\displaystyle x} . Your δx{\displaystyle \delta x} is .1 (This is the "change" in x{\displaystyle x} from the approximation point to the point you chose.)

3: Take the derivative of your function.

dydx=2x{\displaystyle {\frac {dy}{dx}}=2x} . Now, "split" this up (This is not really what is happening, but to keep things simple, assume you are "multiplying" dx{\displaystyle dx} over.)

3b. Now you have dy=2x⋅dx{\displaystyle dy=2x\cdot dx} . We are assuming dy,dx{\displaystyle dy,dx} are approximately the same as the change in x{\displaystyle x} , thus we can use Δx{\displaystyle \Delta x} and y{\displaystyle y} .

4: To find F(4.1){\displaystyle F(4.1)} , take F(4)+dy{\displaystyle F(4)+dy} to get an approximation. 16 + 0.8 = 16.8; This approximation is nearly exact (The real answer is 16.81. This is only one hundredth off!)

If f(x),g(x){\displaystyle f(x),g(x)} are continuous on the closed interval [a,b]{\displaystyle [a,b]} and differentiable on the open interval (a,b){\displaystyle (a,b)} , g(a)≠g(b){\displaystyle g(a)\neq g(b)} and g′(x)≠0{\displaystyle g'(x)\neq 0} , then there exists a number c∈(a,b){\displaystyle c\in (a,b)} such that